# [69. Sqrt(x)](https://leetcode.com/problems/sqrtx/description/)
# 思路
求int型数的平方根并向下取整。   
考虑采用二分法。    
注意：为了防止溢出，判断mid * mid与 x 的值时用除法，即判断 mid 与 x / mid的大小关系。   
# C++
``` C++
class Solution {
public:
    int mySqrt(int x) {
        if(x < 2) return x;
        int low = 1, high = x / 2, mid;
        while(low <= high){
            mid = (low + high) / 2;
            if(mid == x / mid) return mid;
            else if(mid < x / mid) low = mid + 1;
            else high = mid - 1;
        }
        return high;  
    }
};
```
